Download ppt - Gentle Introduction to Operating System

Gentle Introduction to Operating System

Anthony K. H. Tung(鄧锦浩）Associate Professor

Department of Computer Science

http://www.comp.nus.edu.sg/~atunghttp://www.renren.com/profile.do?id=313870900&_ua_flag=93w

Objectives Operating System in SOC is a second year

course(CS2106) which have prerequisites of either CS1104/ CS2100 : Computer Organization EE2007: Microprocessor Systems

Aim of this talk: Provide some hardware background so that you

can understand OS better Explain “why” instead of just “how” and

“what”? Provide a framework that can merge different

concepts together

Approach We will design an OS together, starting from

a simple one and finally to a complex one: Single user, single process Single user, multiple processes Multiple users, multiple processes

While designing, we will take into consideration various desiderata Efficiency Robustness to changes Robustness to errors Consistency Isolation

What is OS?

As the name implied, “operating system” is a software system that operate something? What is that “something”?

“something”=hardware Why do we need to put in a software

layer that operate the hardware? To appreciate something, we need to

ask ourselves what is something do not exist…

In the beginning…

There is no such thing as an OS, each user write their own program to control the hardware which have their own control code or instruction set

Example: hard disk Have many tracks Each track have many

sectors Data are stored in each

sector as binary bits1010010100….

Eg. of Instruction Set for Hard Disk

Physically the disk is control by electronic circuit

instruction sector no. memory loc. 1 0 0 1 0 1 1 1 0 1 1 0

Code Description Binary

00 Reset Disk Drives 00

01 Check Drive Status 01

02 Read Sectors From Drive 10

03 Write Sectors To Drive 11

Storing Data as Files When there was no OS(i.e. no fgets,

fopen…), programmers have to organize the sectors into files in their programs

Example: Use the first sector as file directory and indicate what are the sectors use to store data for each file. Figure on the right show file directory for two files, “numbers” and “letters”

“numbers”, 2, 5, 8, -1“letters”, 3, 6, 1023, -1 . . .

20, 4, 100, …………

A, Z, B, H, ……….

900, 1000, 50, …………

B, C, I, A, ……….

54, 20, 10,………….

O, R, Q, G …………

Sector 2

Sector 3

Sector 4

Sector 5

Sector 6

Sector 7

Sector 8

Sector 9

Sector 1023

Sector 1024

Sector 1

filename

End marker

File directory sector

.

.

Storing Data as Files(II)

How to read the file “numbers”?

“numbers”, 2, 5, 8, -1“letters”, 3, 6, 1023, -1 . . .

20, 4, 100, …………

A, Z, B, H, ……….

900, 1000, 50, …………

B, C, I, A, ……….

54, 20, 10,………….

O, R, Q, G …………

Sector 2

Sector 3

Sector 4

Sector 5

Sector 6

Sector 7

Sector 8

Sector 9

Sector 1023

Sector 1024

Sector 1

filename

End marker

Step 1: 10, Sector 1, Mem. Location 20 // Read Sector 1 // into memory location 20Step 2: Search location 20 for filename=“numbers”;Step 3: sector_num = first sector of “numbers”;Step 3: While (sector_num<>-1) doStep 4: { 10, Sector sector_num, Mem. Location 21;Step 5: process data in mem. Location 21;Step 6: }

Note: “10” is the hardware code for reading a sector of the disk

.

.

What is the problem of such an approach?

Difficulty in programming. Have to remember the command code and repeat the same thing in every program.

Different programmers might implement the file differently. What if one of them resign?

Not robust to change: What if a different brand of disk with different instruction code is used? i.e. “01” is used to read a sector instead?

Having an OS avoid these problems

In its most basic form, OS provide a list of routines or services that are called by application programs to perform certain tasks through software interrupts

Details of such call involve three concepts: Program Counter (PC) Interrupt Vector Table

(IVT) Context Switch

Open File

Close File

Get System Data

Get/Set File Attribute

Read file

Write file

Allocate Memory

Write to Screen

Write to Printer

. .For (i=0;i<10;i++) write file(“numbers”,…) . . .

0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1000-2000

OS Services

ApplicationProgram

Program Counter (PC)

The program counter(PC) is a register that store the address of the next instruction to be executed by the CPU

int i=5, j=6, k=7;

i=i+1;

j=j+2;

k=i+j;

i=0;

end

1000

1002

1003

1004

1005

1006

1007

1008

Memory address

1001

Program

variables

ij

kPC: 1000

CPU:

At the start, PC point to start of program



int i=5, j=6, k=7;

i=i+1;

j=j+2;

k=i+j

i=0;

end

5

6

7

1000

1002

1003

1004

1005

1006

1007

1008

Memory address

1001

Program

variables

ij

kPC: 1001

CPU:int i=5, j=6, k=7;

Fetch instruction at 1000 and execute. Increment PC to 1001



int i=5, j=6, k=7;

i=i+1;

j=j+2;

k=i+j

i=0;

end

6

6

7

1000

1002

1003

1004

1005

1006

1007

1008

Memory address

1001

Program

variables

ij

kPC: 1002

CPU:i=i+1




int i=5, j=6, k=7;

i=i+1;

j=j+2;

k=i+j

i=0;

end

6

8

7

1000

1002

1003

1004

1005

1006

1007

1008

Memory address

1001

Program

variables

ij

kPC: 1003

CPU:j=j+2




int i=5, j=6, k=7;

i=i+1;

j=j+2;

k=i+j

i=0;

end

0

8

14

1000

1002

1003

1004

1005

1006

1007

1008

Memory address

1001

Program

variables

ij

kPC: 1005

CPU:k=i+j;

Fetch instruction at 1004 and execute. Increment PC to 1005 and end program

Interrupt Vector Table(IVT)

During interrupt, the program counter(PC) must be set to point to the address of the OS service in order to execute the routine there. How do the program know where is the address?

IVT: store the location of each interrupt service

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer

. .For (i=0;i<10;i++) write file(“numbers”,…) . . .

0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1000-2000

OS Services

ApplicationProgram

PC: 0043

CPU:Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021

04 Get/Set File Attribute

0029

05 Read file 0036

06 Write file 0043

07 ..

Interrupt Vector Table

Let see how a software interrupt happen now! (User Mode)

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer

INT 01, “numbers”

0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 1002

CPU:

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


Let say the application program is running and the PC is pointing at location 1002

i=i+1;1003


Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 1003

CPU:INT 01, “numbers”

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


It load in the instruction and see that it is an interrupt command(INT) asking for interrupt service 01

i=i+1;1003


Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 1003


Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


It look up the IVT and see that the service “Open File” is locate at address 0001

i=i+1;1003

Let see how a software interrupt happen now! (Kernel Mode)

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 0001

CPU:

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


It then set PC to 0001 and start fetching instructions from the “Open File” service there

i=i+1;1003

Let see how a software interrupt happen now! (Kernel Mode)

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 0005

CPU: End of Open File

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


Assuming we reach the end of “Open File” routine, then how do we know where we should continue next?

i=i+1;1003

Context(I): User Mode

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 1003


Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043

Interrupt Vector TableActually, when an interrupt happens, we need to save the context of the application program before we jump to the interrupt service routine!

i=i+1;1003

Program XXX 1003Context stack

Context(II): Kernel Mode

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 0005

CPU: End of Open File

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


So when read the end of “Open File”, we look at the context stack and jump back to the place where interrupt occurs!

i=i+1;1003


Context(III): return from interrupt(User mode again!)

Open File

Close File

Get System Data


Read file

Write file

Allocate Memory

Write to Screen

Write to Printer


0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

1002

OS Services

ApplicationProgram

PC: 1003

CPU:

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


So when read the end of “Open File”, we look at the context stack and jump back to the place where interrupt occurs!

i=i+1;1003


Context(IV): Final Notes

Note that in a context stack, there can be many entries as if many different processes/programs are running at the same time and issues different interrupts at different point in time

Program XXX 1003

Context stack Process YYY 2500

Process ZZZ 0500

Program AAA 0700

program address

Making OS more efficient

Since the OS provide a set of routines/services that are called by all application programs, it make sense to ensure that these routines/services are efficient so that the whole computer system is efficient as a whole

We will next look at two concepts that make the OS more efficient in using the CPU: Caching Hardware interrupts

Caching(Motivation)

CPUDisk

We wish to read 100 numbers from the disk and sum them up. These are the time that is need for each operations:

Initialize hard disk: 5msRead 1 number: 2 msTransfer 1 number: 2 msSum 2 numbers: 0.5ms

How long will it take to finish what we want to do?

Initialize hard disk: 5ms

Read 1 number: 2 ms

Transfer 1 number: 2 ms

Sum 2 numbers: 0.5 ms

Caching(Motivation)

CPUDisk

Assuming no cache/bufferInitialize 100 times for 100 numbers: 5 x 100 ms = 500 msRead 100 numbers: 2 x 100ms = 200msTransfer 100 numbers: 2 x 100ms = 200msSum 100 numbers: 100 x 0.5 = 50msTotal = 950ms


Read 1 number: 2 ms



Caching(Motivation)

CPU

Disk

Assuming we DO have cache that can store all 100 numbersInitialize 1 times for 100 numbers: 5 x 1 ms = 5 msRead 100 numbers: 2 x 100ms = 200msTransfer 100 numbers: 2 x 100ms = 200msSum 100 numbers: 100 x 0.5 = 50msTotal = 455ms


Read 1 number: 2 ms



100, 29, 50, 200,…, 90

Local buffer of 100 numbers

100, 29, 50, 200,…, 90

CPU Memory Cache

Caching(Motivation)

CPU

Disk

Because of this, data on hard disk are stored as a block called “sector”. One sector can contain one set of numbers. Only the first read of the sector will cause access to the disk. If the sector is stored in the memory cache, subsequent read will be from the cache.


Read 1 number: 2 ms



100, 29, 50, 200,…, 90


100, 29, 50, 200,…, 90

CPU Memory Cache

Caching(Motivation)

CPU

Disk

Can we do better? If we are summing N numbers, does it always make sense to have a local buffer or sector of N numbers for the disk?

How about first loading 50 numbers, let the CPU start to compute and then CURRENTLY load the other 50 numbers?


Read 1 number: 2 ms



100, 29, 50, 200,…, 90


100, 29, 50, 200,…, 90

CPU Memory Cache

Caching & Interrupt (Motivation)

Concurrent reading and calculating

CPU

Initialize hard disk: 5msRead 1 number: 2 msTransfer 1 number: 2 msSum 2 numbers: 0.5ms

Disk

(205ms)idle

Initialize (5ms)

hardware disk interrupt

Read 50 No. (100ms)

Transfer 50 No. (100ms)

Initialize command

( 25ms)Sum first 50 No.

Initialize 2 (5ms)

Harkware disk interrupt

Read 50 No. (100ms)

Initialize command 2


( 25ms)Sum next 50 No.

(180ms)idle

idle (25ms)

Total time = 205ms + 205ms + 25ms = 435ms

Time is saved since summing the first 50 numbers is done concurrent with disk I/O

What if we divide the numbers into 4 blocks of 25 numbers ? Optimal number of blocks?

Interrupt and Currency The concurrent execution in previous example is

possible because of hardware interrupt AND caching. The CPU leave the task of reading data to the hard

disk and perform other task(adding number). The disk inform the CPU using interrupt when it complete its reading and transferring of data

Notice that we have more from single program, single process to single program with 2 processes (multi-threaded)

Total = 0;For (i=0; i<100;i++) number[i]=-1;

For (i=0; i<100;i++) fscanf(input,"%d",&number[i]);

For (i=0; i<100;i++){ while number[i]==-1; Total=Total+number[i]);}

Fork()

producer

consumer

Memory Hierarchy for Caching

In general, CPU only read data from the processor registers and CPU cache

Cache miss result in fetching from the next layer

Pipelining(流水线 ) In general, caching

and buffering serve to coordinate devices with different speed

Who handle hardware interrup? Same as in software

interrupt! Hardware interrupt sent

signal to CPU which will then give control to one of the OS interrupt handling routine

Open File

Close File

Get System Data


Read file

Write file

Handle disk interrupt

Scheduler

Write to Printer

0001-005

0021-0028

0029-0035

0036-0042

0043-0055

0056-0068

0069-0080

0036-0042

Memory address

0006-0020

Interrupt No.

Service Address

01 Open File 0001

02 Close File 0006

03 Get SD 0021


0029

05 Read file 0036

06 Write file 0043


Single User, Multiple Programs

CPU

Disk

(205ms)idle

Initialize (5ms)

hardware disk interrupt

Read 50 No. (100ms)


Initialize command

( 25ms)Sum first 50 No.

Initialize 2 (5ms)

Harkware disk interrupt

Read 50 No. (100ms)

Initialize command 2


( 25ms)Sum next 50 No.

(180ms)idle

idle (25ms)

Notice that the CPU still have idle time in the ADD_100_NUMBER example earlier. What can we do with these idle time?

Answer: Swap the context of other programs in and run (Just like in Windows)

Clock interrupt

A hardware implemented clock issue a clock interrupt at regular interval. Eg. intel process give a clock interrupt to the OS every 15ms. This can be changed by the OS which can use this clock as an “alarm clock”

This “alarm clock” is used to wake up a Scheduler which will select one of the programs to be ran in the CPU

Multitasking/Time Sharing Using Clock

Open File

Close File

Scheduler


.

.

.

Process XXX

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

1000-2000PC: 1003

CPU:Instruction from XXX

Program YYY2001-2800

Context stack Program YYY 2200

Process ZZZ 3010

Process ZZZ2801-3500

CPU initially running Process XXX at location 1003

Clock Interrupt Occur!


Open File

Close File

Scheduler


.

.

.

Process XXX

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

1000-2000PC: 0021

CPU:Scheduler



Process ZZZ 3010


Context of Process XXX saved and Scheduler loaded

Process XXX 1003


Open File

Close File

Scheduler


.

.

.

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

PC: 0021

CPU:Scheduler


Process ZZZ 3010

Scheduler select Program YYY to be ran next

Process XXX 1003Process XXX1000-2000




Open File

Close File

Scheduler


.

.

.

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

PC: 2200

CPU:Instruction from YYY

Context stack Process ZZZ 3010

Scheduler select Program YYY to be ran next

Process XXX 1003

Process XXX1000-2000




Open File

Close File

Scheduler


.

.

.

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

PC: 2600

CPU:Instruction from YYY

Context stack Process ZZZ 3010

Then after a certain period, clock interrupt occur again

Process XXX 1003





Open File

Close File

Scheduler


.

.

.

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

PC: 0021

CPU:Scheduler

Context stack

Process ZZZ 3010

Context of Programm YYY saved

Scheduler is loaded again

Process XXX 1003

Program YYY 2600





Open File

Close File

Scheduler


.

.

.

Process XXX

0001-005

0021-0028

0029-0035

.

..

Memory address

0006-0020

1000-2000PC: 3010

CPU:Instruction from ZZZ


Context stack


Scheduler decide to load Process ZZZ next

Process XXX 1003

Program YYY 2600

Multitasking/Time Sharing

Note that sometimes program are swapped out by the scheduler not because its time slice is used up but because it is waiting for some input/output (eg. reading disk) and is thus idle

This is made known to the OS because input/output are also handled by the OS through software interrupts!

Isolation: Security and Protection Since multiple programs are at

different stage of their execution concurrently in the system, care must be take care that they are isolated from each other i.e. they don’t affect each other through writing/reading from each others memory or file.

Again, this can be control by the OS since all these are done through its interrupt services

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $100

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

Bank Data Yapeng Deposit $50

Read Record(Yapeng, BC345, $100)

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $150

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

Bank Data Yapeng Deposit $50


Write Record(Yapeng, BC345, $150)

Very simple?

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $150

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

Data Yapeng Deposit $30



Now both Yapeng and his wife Wang Fei deposit money concurrently

Wang Fei Deposit $100

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $250

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

Data Yapeng Deposit $30



Wang Fei update record first



Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $180

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

Database Yapeng Deposit $30



Follow by Yapeng’s update



They lost $100!

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $150

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX



Solution: add a lock, no others can access a record when it is locked

locked

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $150

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX






locked

Data Consistency

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $180

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX

XXX XXX XXXX






locked



Data Deadlock

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $150

XXX XXX XXXX

Wang Fei BC987 $1000

XXX XXX XXXX

XXX XXX XXXX

Database

Read Record (Yapeng, BC345, $150)Read Record(Wang Fei, BC987,$1000)Write Record (Yapeng, BC345, $100)Write Record (Wang Fei, BC345, $1050)


Now Yapeng want to transfer $50 to Wang Fei and Wang Fei want to transfer $100 to Yapenglocked



Read Record(Wang Fei, BC987,$1000)Read Record (Yapeng, BC345, $150)Write Record (Yapeng, BC345, $250)Write Record (Wang Fei, BC345, $900)

Read Record(Wang Fei, BC987, $1000)

locked

Data Deadlock

Name Account No.

Savings

XXX XXX XXXX

XXX XXX XXXX

Yapeng BC345 $150

XXX XXX XXXX

Wang Fei BC987 $1000

XXX XXX XXXX

XXX XXX XXXX

Database

Read Record (Yapeng, BC345, $150)Read Record(Wang Fei, BC987,$1000)Write Record (Yapeng, BC345, $100)Write Record (Wang Fei, BC987, $1050)


Both transactions cannot continue because they are holding each other’s recordlocked



Read Record(Wang Fei, BC987,$1000)Read Record (Yapeng, BC345, $150)Write Record (Yapeng, BC345, $200)Write Record (Wang Fei, BC987, $900)

Read Record(Wang Fei, BC987, $1000)

locked

Memory & Process Management

Memory is an important resource since processes need to be load into memory to be ran and there might not be enough memory to store all the processes and their data

Thus, memory and process management come hand in hand

Processes/programs are also a type of “data” to the OS

Open File

Close File

Scheduler


.

.

.

0001-005

0021-0028

0029-0035

Memory address

0006-0020




Memory & Process Management

Imagine Program YYY is the following program:

Open File

Close File

Scheduler


.

.

.

0001-005

0021-0028

0029-0035

Memory address

0006-0020




main(){ for (int i=2;i<100;i++) { if IsPrime(i) printf(“%i is prime”,i); }}

Boolean IsPrime(int N){ int IsPrime=1; for (int i=2;i<N;i++) { if (N mod i==0) return FALSE } return TRUE}

How do main() know the location of IsPrime()? Essentially set by OS when it load Program YYY. Program YYY might be moved to different location

Relative Address: Call function that is 10 locations behind

Virtual Memory

Virtual Memory is analogy to virtual money

The bank told you that you have $10,000 with them but You do not know the

physical location The bank might stored

$10,000 gold bar and not dollar notes

In fact the bank might not have so much money

Virtual Memory vs File Management

Virtual Memory File Management

Make use of disk to store more processes/data

Make use of disk to store more data

Transparent to the processes

Processes are aware of the existence of file

Need special hardware (MMU) for implementation

Just make use of ram and disk. Handle by OS.

Need to swap in/out both processes and data

Need to swap in/out only the data

Many concepts are however similar and have the same reasoning

Virtual Memory

Each process is divided into pages (similar to sector/block for disk/file) and bring into the memory as a whole unit. Assumption of locality:

When an instruction(data) in a page (block) is accessed, other instructions (data) in the same page is likely to be accessed

Need to monitor each virtual page is mapped to which physical frame But this is also in the

main memory which require memory access!

Virtual Memory (Paging hardware: TLB)

Translation look-aside buffers (TLBs) Similar to the concept of caching. Part of the page table is cached in the TLB which have faster access

time than main memory RAM If page (block) not in memory(invalid bit set), issue a page fault through hardware interrupt (by MMU). An

interrupt handler from the OS will then be activated to do the swapping

Virtual Memory(Page Replacement)

If there is no free frame in the physical memory, need to move a page to the disk.

Page Replacement Policy FIFO(First In, First Out): Throw out the

earliest page that came in. Need to monitor the time a page is brought in

LRU(Least Recently Used): Throw out the page that last access time is furthest away. Need to monitor the time a page is last access

Virtual Memory(Thrashing)

Example: A process XXX is given a time slice of 5 clock tick/interrupts to be executed in the CPU. Assuming 1 clock interrupt occur every 15ms, this mean XXX is given 15*5=75ms to use the CPU.

Assuming page fault for XXX and it take 100ms to load in the page for XXX. Then XXX has not enough time to be executed! Same thing will happen the next time process XXX is scheduled to be ran.

Happened when many processes is in the system and each process get little time slices and memory

Summary: Looking back

Why do we rely on OS to manage hardware, resources etc.? Answer: Because everything go through the OS in the

form of hardware and software interrupts and this make OS the center of the universe

Relook at the important concepts Software Interrupts Caching(for data) Hardware Interrupts

Disk interrupt Clock interrupt (Time Slicing) MMU interrupt (Page Fault)

Virtual Memory Paging for both process and data

Question to think about

Software interrupts seems to be similar to function call in your C programming. What is the different between interrupt handler and function in C?

Useful Reading and References

History of operating system http://en.wikipedia.org/wiki/History

_of_operating_systems Interrupts in DOS

http://www.8bs.com/othrdnld/manuals/512/512tech/m512techa.htm

Questions and Contact

[email protected] http://www.comp.nus.edu.sg/

~atung http://www.renren.com/akhtung